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目的:鉴定1例NS5B区和E1区PCR产物测序结果分型不一致的HCV病毒株为基因重组还是混合感染。方法:提取血浆的病毒总RNA,逆转录后进行巢式PCR扩增,以NS5B和E1编码区为目的基因进行PCR扩增并测序,通过构建进化树进行基因分型,发现两个编码区基因分型的结果不同,对NS5B和E1区进行克隆测序,每个编码区分别挑取50个克隆进行测序和基因分型;针对HCV全长序列分别设计7对型特异性引物并进行PCR扩增。结果:NS5B(8 256~8 641 bp)区PCR产物测序后的分型结果为1b,E1(697~1 322 bp)区为2a;NS5B(8 266~9 274 bp)基因的50个克隆测序结果均为1b,C/E1(849~2 152 bp)基因的50个克隆测序结果均为2a;7个片段的PCR型特异性扩增以及进化分析发现,在同一区域既存在1b亚型,也存在2a亚型。结论:该病例为1b和2a的混合感染。
OBJECTIVE: To identify whether one of the HCV genotypes in which NS5B region and E1 region PCR products were sequenced inconsistently was gene recombinant or mixed. Methods: Total RNA was extracted from plasma and reverse-transcribed for nested PCR. The NS5B and E1 coding regions were amplified by PCR and sequenced. The two genomic regions were identified by phylogenetic tree analysis The results of typing were different. The NS5B and E1 regions were cloned and sequenced, and 50 clones were selected for sequencing and genotyping in each coding region. Seven pairs of specific primers were designed and PCR amplified . RESULTS: The genotyping results of PCR products of NS5B (8 256 ~ 8 641 bp) were 1b and E1 (697 ~ 1 322 bp) was 2a. Fifty clones of NS5B (8 266 ~ 9 274 bp) The results showed that all of the 50 clones of the C / E1 (849-2 152 bp) gene were 2a. The PCR amplification of seven fragments and the evolutionary analysis showed that there were two subtypes 1b in the same region, There are also 2a subtypes. Conclusion: This case is a mixed infection of 1b and 2a.